Am 08.07.2013 15:39, schrieb John Kliewe:
Thanks very much Urs -- the tip about moving the \override outside of
the \times did the trick.
The reason I duplicated the entire phrase in each voice is that I want
to see the second group of 16ths in a single-beam group. By limiting
the dual-voice treatment only to the small tuple that really requires
it, I am not able to join the beams the way it appears in the original
Paderewski edition that I am typesetting. My example has the beams I
want, but the slur is much too high. Your suggestion corrects that
high slur, but doesn't join the beam the way I want.
Obviously I didn't look close enough. I was so focused on the high slur
that I didn't realize the difference in beaming :-[
Is there a way to accomplish both? I know that both versions are
technically correct, but I would like to mimic the original if possible.
Yes, there is.
Your problem is that the b es, ais triplet isn't in the same voice
context as the preceding 16th notes, therefore they can't share the beam
(as they couldn't share a slur either for example).
So what you need to do is get these notes in the same voice .
There are two steps involved:
1) Change the polyphonic construct:
<<
{ [music one] }
\\
{ [music two] }
creates two new Voice contexts whereas
<<
{ [music one] }
\new Voice {
[music two]
}
only creates a new Voice context for 'music two' while 'music one'
seamlessly continues the Voice context that has been in effect before
(and afterwards.
So what you want is to have the triplets that you want to beam in the
place of 'music one'. So:
2) Switch the expressions:
Try replacing the triplet section with
<<
{
\voiceTwo
\times 2/3 {
{
b16 eis, ais
}
}
}
\new Voice
{
\once \override TupletNumber #'stencil = ##f
\times 2/3 {
b16^\accent[ s ais]
}
}
>>
and it should work.
Notice the \voiceTwo which is necessary now to switch the stems
downward, and the circumflex to place the accent above the voice.
HTH
Urs
--John K
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